How to calculate $\int xe^{1/x^2} \ dx$ How would I go about integrating $\int xe^{1/x^2} \ dx$? 
I attempted a u-substitution and integration by parts already, but that didn't get me anywhere.
 A: Even if the result is not the most simple, it corresponds to a closed form (even if the final result does not involve only elementary terms). 
You want to compute
$$I=\int x\, e^{\frac{1}{x^2}}\,dx$$ Let us perform one integration by parts $$du=x\,dx \implies u=\frac{x^2}{2}$$ $$v=e^{\frac{1}{x^2}}\implies dv=-\frac{2 e^{\frac{1}{x^2}}}{x^3}\, dx$$ So $$I=\int x\, e^{\frac{1}{x^2}}\,dx=\frac{1}{2} e^{\frac{1}{x^2}} x^2+\int\frac{e^{\frac{1}{x^2}}}{x}\,dx$$ Now, consider $$J=\int\frac{e^{\frac{1}{x^2}}}{x}\,dx$$ and change variable $x=\frac{1}{\sqrt{y}}$, $dx=-\frac{1}{2 y^{3/2}}\,dy$.
So $$J=\int\frac{e^{\frac{1}{x^2}}}{x}\,dx=-\frac 12 \int\frac{e^y}{ y}\,dy$$ The last integral is the definition of $\text{Ei}(y)$, the exponential integral. Back to $x$, we then get the result given by Wolfram Alpha $$I=\frac{1}{2} \left(e^{\frac{1}{x^2}} x^2-\text{Ei}\left(\frac{1}{x^2}\right)\right)$$ If you have to compute the value of this integral between bounds, for large values of $x$, you could use as an approximation $$\text{Ei}\left(\frac{1}{x^2}\right)=\gamma-2\log(x)
  +\frac{1}{x^2}+\frac{1}{4 x^4}+\frac{1}{18 x^6}+\frac{1}{96
   x^8}+O\left(\left(\frac{1}{x}\right)^9\right)$$
A: $\int xe^\frac{1}{x^2}~dx$
$=\int x\sum\limits_{n=0}^\infty\dfrac{x^{-2n}}{n!}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{x^{1-2n}}{n!}dx$
$=\int\left(x+\dfrac{1}{x}+\sum\limits_{n=2}^\infty\dfrac{x^{1-2n}}{n!}\right)dx$
$=\int\left(x+\dfrac{1}{x}+\sum\limits_{n=0}^\infty\dfrac{x^{-2n-3}}{(n+2)!}\right)dx$
$=\dfrac{x^2}{2}+\ln x+\sum\limits_{n=0}^\infty\dfrac{x^{-2n-2}}{(n+2)!(-2n-2)}+C$
$=\ln x+\dfrac{x^2}{2}-\sum\limits_{n=0}^\infty\dfrac{1}{2(n+2)!(n+1)x^{2n+2}}+C$
